The number of 3-digit numbers,formed using th | vedclass

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(D) The given digits are $S = \{2, 3, 4, 5, 7\}$. The total number of $3$-digit numbers that can be formed without repetition is $^5P_3 = 5 	imes 4 \

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(D) The given digits are $S = \{2, 3, 4, 5, 7\}$. The total number of $3$-digit numbers that can be formed without repetition is $^5P_3 = 5 	imes 4 	imes 3 = 60$.$A$ number is divisible by $3$ if the sum of its digits is divisible by $3$.We need to find combinations of $3$ digits from $S$ whose sum is divisible by $3$:$1. \{2, 3, 4\} ightarrow 	ext{sum} = 9$ (divisible by $3$)$2. \{2, 4, 3\}$ (same as above)$3. \{3, 4, 5\} ightarrow 	ext{sum} = 12$ (divisible by $3$)$4. \{5, 7, 3\} ightarrow 	ext{sum} = 15$ (divisible by $3$)$5. \{7, 2, 3\} ightarrow 	ext{sum} = 12$ (divisible by $3$)$6. \{4, 5, 3\}$ (same as above)Let's list all subsets of size $3$ and their sums:- $\{2, 3, 4\} = 9$ (Divisible)- $\{2, 3, 5\} = 10$- $\{2, 3, 7\} = 12$ (Divisible)- $\{2, 4, 5\} = 11$- $\{2, 4, 7\} = 13$- $\{2, 5, 7\} = 14$- $\{3, 4, 5\} = 12$ (Divisible)- $\{3, 4, 7\} = 14$- $\{3, 5, 7\} = 15$ (Divisible)- $\{4, 5, 7\} = 16$There are $4$ sets of digits whose sum is divisible by $3$: $\{2, 3, 4\}, \{2, 3, 7\}, \{3, 4, 5\}, \{3, 5, 7\}$.Each set can form $3! = 6$ numbers.Total numbers divisible by $3 = 4 	imes 6 = 24$.Total numbers not divisible by $3 = 60 - 24 = 36$.

The number of $3$-digit numbers,formed using the digits $2, 3, 4, 5$ and $7$,when the repetition of digits is not allowed,and which are not divisible by $3$,is equal to ..........

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